\(\int \frac {x^2 (A+B x^2+C x^4)}{\sqrt [3]{a+b x^4}} \, dx\) [99]

Optimal result

Integrand size = 27, antiderivative size = 138 \[ \int \frac {x^2 \left (A+B x^2+C x^4\right )}{\sqrt [3]{a+b x^4}} \, dx=\frac {3 C x^3 \left (a+b x^4\right )^{2/3}}{17 b}+\frac {(17 A b-9 a C) x^3 \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{51 b \sqrt [3]{a+b x^4}}+\frac {B x^5 \sqrt [3]{1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{4},\frac {9}{4},-\frac {b x^4}{a}\right )}{5 \sqrt [3]{a+b x^4}} \] Output:

3/17*C*x^3*(b*x^4+a)^(2/3)/b+1/51*(17*A*b-9*C*a)*x^3*(1+b*x^4/a)^(1/3)*hyp 
ergeom([1/3, 3/4],[7/4],-b*x^4/a)/b/(b*x^4+a)^(1/3)+1/5*B*x^5*(1+b*x^4/a)^ 
(1/3)*hypergeom([1/3, 5/4],[9/4],-b*x^4/a)/(b*x^4+a)^(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 10.08 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.76 \[ \int \frac {x^2 \left (A+B x^2+C x^4\right )}{\sqrt [3]{a+b x^4}} \, dx=\frac {\sqrt [3]{1+\frac {b x^4}{a}} \left (35 A x^3 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )+21 B x^5 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{4},\frac {9}{4},-\frac {b x^4}{a}\right )+15 C x^7 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {7}{4},\frac {11}{4},-\frac {b x^4}{a}\right )\right )}{105 \sqrt [3]{a+b x^4}} \] Input:

Integrate[(x^2*(A + B*x^2 + C*x^4))/(a + b*x^4)^(1/3),x]
 

Output:

((1 + (b*x^4)/a)^(1/3)*(35*A*x^3*Hypergeometric2F1[1/3, 3/4, 7/4, -((b*x^4 
)/a)] + 21*B*x^5*Hypergeometric2F1[1/3, 5/4, 9/4, -((b*x^4)/a)] + 15*C*x^7 
*Hypergeometric2F1[1/3, 7/4, 11/4, -((b*x^4)/a)]))/(105*(a + b*x^4)^(1/3))
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.14, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2432, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^2 \left (A+B x^2+C x^4\right )}{\sqrt [3]{a+b x^4}} \, dx\)

\(\Big \downarrow \) 2432

\(\displaystyle \int \left (\frac {A x^2}{\sqrt [3]{a+b x^4}}+\frac {B x^4}{\sqrt [3]{a+b x^4}}+\frac {C x^6}{\sqrt [3]{a+b x^4}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {A x^3 \sqrt [3]{\frac {b x^4}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )}{3 \sqrt [3]{a+b x^4}}+\frac {B x^5 \sqrt [3]{\frac {b x^4}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {5}{4},\frac {9}{4},-\frac {b x^4}{a}\right )}{5 \sqrt [3]{a+b x^4}}+\frac {C x^7 \sqrt [3]{\frac {b x^4}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {7}{4},\frac {11}{4},-\frac {b x^4}{a}\right )}{7 \sqrt [3]{a+b x^4}}\)

Input:

Int[(x^2*(A + B*x^2 + C*x^4))/(a + b*x^4)^(1/3),x]
 

Output:

(A*x^3*(1 + (b*x^4)/a)^(1/3)*Hypergeometric2F1[1/3, 3/4, 7/4, -((b*x^4)/a) 
])/(3*(a + b*x^4)^(1/3)) + (B*x^5*(1 + (b*x^4)/a)^(1/3)*Hypergeometric2F1[ 
1/3, 5/4, 9/4, -((b*x^4)/a)])/(5*(a + b*x^4)^(1/3)) + (C*x^7*(1 + (b*x^4)/ 
a)^(1/3)*Hypergeometric2F1[1/3, 7/4, 11/4, -((b*x^4)/a)])/(7*(a + b*x^4)^( 
1/3))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2432

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[ 
Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n, p}, x] && (PolyQ[Pq, x] || Poly 
Q[Pq, x^n])
 

Maple [F]

\[\int \frac {x^{2} \left (C \,x^{4}+B \,x^{2}+A \right )}{\left (b \,x^{4}+a \right )^{\frac {1}{3}}}d x\]

Input:

int(x^2*(C*x^4+B*x^2+A)/(b*x^4+a)^(1/3),x)
 

Output:

int(x^2*(C*x^4+B*x^2+A)/(b*x^4+a)^(1/3),x)
 

Fricas [F]

\[ \int \frac {x^2 \left (A+B x^2+C x^4\right )}{\sqrt [3]{a+b x^4}} \, dx=\int { \frac {{\left (C x^{4} + B x^{2} + A\right )} x^{2}}{{\left (b x^{4} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(x^2*(C*x^4+B*x^2+A)/(b*x^4+a)^(1/3),x, algorithm="fricas")
 

Output:

integral((C*x^6 + B*x^4 + A*x^2)/(b*x^4 + a)^(1/3), x)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.54 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.88 \[ \int \frac {x^2 \left (A+B x^2+C x^4\right )}{\sqrt [3]{a+b x^4}} \, dx=\frac {A x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [3]{a} \Gamma \left (\frac {7}{4}\right )} + \frac {B x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [3]{a} \Gamma \left (\frac {9}{4}\right )} + \frac {C x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt [3]{a} \Gamma \left (\frac {11}{4}\right )} \] Input:

integrate(x**2*(C*x**4+B*x**2+A)/(b*x**4+a)**(1/3),x)
 

Output:

A*x**3*gamma(3/4)*hyper((1/3, 3/4), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*a 
**(1/3)*gamma(7/4)) + B*x**5*gamma(5/4)*hyper((1/3, 5/4), (9/4,), b*x**4*e 
xp_polar(I*pi)/a)/(4*a**(1/3)*gamma(9/4)) + C*x**7*gamma(7/4)*hyper((1/3, 
7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(1/3)*gamma(11/4))
 

Maxima [F]

\[ \int \frac {x^2 \left (A+B x^2+C x^4\right )}{\sqrt [3]{a+b x^4}} \, dx=\int { \frac {{\left (C x^{4} + B x^{2} + A\right )} x^{2}}{{\left (b x^{4} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(x^2*(C*x^4+B*x^2+A)/(b*x^4+a)^(1/3),x, algorithm="maxima")
 

Output:

integrate((C*x^4 + B*x^2 + A)*x^2/(b*x^4 + a)^(1/3), x)
 

Giac [F]

\[ \int \frac {x^2 \left (A+B x^2+C x^4\right )}{\sqrt [3]{a+b x^4}} \, dx=\int { \frac {{\left (C x^{4} + B x^{2} + A\right )} x^{2}}{{\left (b x^{4} + a\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(x^2*(C*x^4+B*x^2+A)/(b*x^4+a)^(1/3),x, algorithm="giac")
 

Output:

integrate((C*x^4 + B*x^2 + A)*x^2/(b*x^4 + a)^(1/3), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \left (A+B x^2+C x^4\right )}{\sqrt [3]{a+b x^4}} \, dx=\int \frac {x^2\,\left (C\,x^4+B\,x^2+A\right )}{{\left (b\,x^4+a\right )}^{1/3}} \,d x \] Input:

int((x^2*(A + B*x^2 + C*x^4))/(a + b*x^4)^(1/3),x)
 

Output:

int((x^2*(A + B*x^2 + C*x^4))/(a + b*x^4)^(1/3), x)
 

Reduce [F]

\[ \int \frac {x^2 \left (A+B x^2+C x^4\right )}{\sqrt [3]{a+b x^4}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{4}+a \right )^{\frac {1}{3}}}d x \right ) c +\left (\int \frac {x^{4}}{\left (b \,x^{4}+a \right )^{\frac {1}{3}}}d x \right ) b +\left (\int \frac {x^{2}}{\left (b \,x^{4}+a \right )^{\frac {1}{3}}}d x \right ) a \] Input:

int(x^2*(C*x^4+B*x^2+A)/(b*x^4+a)^(1/3),x)
 

Output:

int(x**6/(a + b*x**4)**(1/3),x)*c + int(x**4/(a + b*x**4)**(1/3),x)*b + in 
t(x**2/(a + b*x**4)**(1/3),x)*a